if the points (1,2) , (3,5) and (5,8) are collinear then the locus of the line containing them?
Answers
Answer:
ANSWER
Three points A,B,C are collinear if direction ratios of AB and BC are proportional.
A(2,3,4) and B(−1,−2,1)
Direction ratios =−1−2,−2−3,1−4
=−3,−5,−3
So, a
1
=−3,b
1
=−5,c
1
=−3
B(−1,−2,1) and C(5,8,7)
Direction ratios =5−(−1),8−(−2),7−1
=6,10,6
so, a
2
=6,b
2
=10,c
2
=6
Now,
a
1
a
2
=
−3
6
=−2
b
1
b
2
=
−5
10
=−2
c
1
c
2
=
−3
6
=−2
Since
a
1
a
2
=
b
1
b
2
=
c
1
c
2
=−2
Therefore, A,B,C are collinear.
Answer:
Slope from (1,4) to (3,-2) is (4-(-2)) / (1–3) = 6/(-2) = -3.
Slope from (3,-2) to (-3,16) is (-2–16) / (3-(-3)) = -18/6 = -3.
Since these are equal, the points are colinear (else they would be vertices of a triangle).
The equation of the line through them is: y-4 = (-3)(x-1) which simplifies to y-7=-3x, or if you prefer y+3x = 7.